After experimentation you have determined the following rate equations:
1. Rate = k[H2] [NO]2
2. Rate = k[H2] 2 [O2]
3. Rate = k[H2 ] [NO]
Identify the order of reaction for each of these rate equations. You must show your workings within your submission, and explain your reasoning
For the first rate equation, Rate = k[H2] [NO]2, we can use the method of initial rates to determine the order with respect to both reactants. This involves running experiments where we vary the initial concentrations of H2 and NO separately, while keeping the other constant, and measuring the initial rate of reaction in each case.
For example, if we run experiments with [H2] = 0.1 M and varying [NO], and measure initial rates of 0.1 M/s, 0.4 M/s, and 0.9 M/s for [NO] = 0.1 M, 0.2 M, and 0.3 M respectively, we can use the following relationship to determine the order with respect to NO:
Rate / [H2] = k[NO]2
At the lowest [NO] tested (0.1 M), we have:
0.1 M/s / 0.1 M = k(0.1 M)2
k = 100 M-2 s-1
At the highest [NO] tested (0.3 M), we have:
0.9 M/s / 0.1 M = k(0.3 M)2
k = 90 M-2 s-1
As these values of k are not exactly the same, we cannot assume a simple order of 2 with respect to NO. Instead, we can use the following relationship to determine the overall order of reaction:
Rate / [H2][NO]2 = k
If we substitute in our values for k and [H2], we get:
Rate / (0.1 M)([NO]2) = 100 M-2 s-1
Rate = 100 M-2 s-1 [H2] [NO]2
This shows that the overall order of reaction is 3, with individual orders of 1 with respect to H2 and 2 with respect to NO.
For the second rate equation, Rate = k[H2]2[O2], we can use the method of initial rates again, but this time with varying concentrations of H2 and O2, while keeping the other constant. For example, if we run experiments with [H2] = 0.1 M and varying [O2], and measure initial rates of 0.1 M/s, 0.2 M/s, and 0.4 M/s for [O2] = 0.1 M, 0.2 M, and 0.3 M respectively, we can use the following relationship to determine the order with respect to O2:
Rate / [H2]2 = k[O2]
At the lowest [O2] tested (0.1 M), we have:
0.1 M/s / (0.1 M)2 = k(0.1 M)
k = 100 M-1 s-1
At the highest [O2] tested (0.3 M), we have:
0.4 M/s / (0.1 M)2 = k(0.3 M)
k = 133.3 M-1 s-1
As these values of k are not exactly the same, we cannot assume a simple order of 1 with respect to O2. Instead, we can use the following relationship to determine the overall order of reaction:
Rate / [H2]2[O2] = k
If we substitute in our values for k and [H2], we get:
Rate = (100 M-1 s-1) [H2]2 [O2]
This shows that the overall order of reaction is 3, with individual orders of 2 with respect to H2 and 1 with respect to O2.
For the third rate equation, Rate = k[H2][NO], we can use the same method as before, but only need to vary the concentration of one reactant while keeping the other constant. For example, if we run experiments with [H2] = 0.1 M and varying [NO], and measure initial rates of 0.1 M/s, 0.2 M/s, and 0.3 M/s for [NO] = 0.1 M, 0.2 M, and 0.3 M respectively, we can use the following relationship to determine the order with respect to NO:
Rate / [H2] = k[NO]
At the lowest [NO] tested (0.1 M), we have:
0.1 M/s / 0.1 M = k(0.1 M)
k = 1 s-1
At the highest [NO] tested (0.3 M), we have:
0.3 M/s / 0.1 M = k(0.3 M)
k = 3 s-1
As these values of k are exactly the same, we can assume a simple order of 1 with respect to NO. We can also see that there is a simple order of 1 with respect to H2, since the rate varies linearly with [H2] in all experiments. Therefore, the overall order of reaction is 2, with individual orders of 1 with respect to both H2 and NO.
To determine the order of reaction for each rate equation, we need to examine the relationship between the rate of reaction and the concentration of reactants.
1. Rate = k[H2] [NO]2:
To determine the order with respect to H2, we can perform experiments while keeping the concentration of NO constant and vary the concentration of H2. Let's consider two experiments:
Experiment 1:
[H2] = x
[NO] = constant
Experiment 2:
[H2] = 2x
[NO] = constant
If we divide the rate of Experiment 2 by the rate of Experiment 1, we get:
(Rate 2) / (Rate 1) = (k[H2] [NO]2) / (k[2H2] [NO]2)
= (k[H2] [NO]2) / (2k[H2] [NO]2)
= 1/2
As the rate is halved when the concentration of H2 is halved, the order of reaction with respect to H2 is 1.
Similarly, if we perform experiments while keeping the concentration of H2 constant and vary the concentration of NO:
Experiment 3:
[H2] = constant
[NO] = y
Experiment 4:
[H2] = constant
[NO] = 2y
(Rate 4) / (Rate 3) = (k[H2] [NO]2) / (k[H2] [(2NO)]2)
= (k[H2] [NO]2) / (4k[H2] [NO]2)
= 1/4
As the rate is one-fourth when the concentration of NO is doubled, the order of reaction with respect to NO is 2.
In summary,
Order of reaction with respect to H2 = 1
Order of reaction with respect to NO = 2
2. Rate = k[H2]2 [O2]:
To determine the order with respect to H2, we can perform experiments while keeping the concentration of O2 constant and vary the concentration of H2. Let's consider two experiments:
Experiment 5:
[H2] = a
[O2] = constant
Experiment 6:
[H2] = 2a
[O2] = constant
(Rate 6) / (Rate 5) = (k[H2]2 [O2]) / (k[(2H2)]2 [O2])
= (k[H2]2 [O2]) / (4k[H2]2 [O2])
= 1/4
As the rate is one-fourth when the concentration of H2 is doubled, the order of reaction with respect to H2 is 2.
Similarly, if we perform experiments while keeping the concentration of H2 constant and vary the concentration of O2:
Experiment 7:
[H2] = constant
[O2] = b
Experiment 8:
[H2] = constant
[O2] = 2b
(Rate 8) / (Rate 7) = (k[H2]2 [O2]) / (k[H2]2 [(2O2)])
= (k[H2]2 [O2]) / (k[H2]2 [4O2])
= 1/4
As the rate is one-fourth when the concentration of O2 is doubled, the order of reaction with respect to O2 is 1.
In summary,
Order of reaction with respect to H2 = 2
Order of reaction with respect to O2 = 1
3. Rate = k[H2 ][NO]:
To determine the order with respect to H2, we can perform experiments while keeping the concentration of NO constant and vary the concentration of H2. Let's consider two experiments:
Experiment 9:
[H2] = c
[NO] = constant
Experiment 10:
[H2] = 2c
[NO] = constant
(Rate 10) / (Rate 9) = (k[H2 ] [NO]) / (k[(2H2)] [NO])
= (k[H2 ] [NO]) / (2k[H2 ] [NO])
= 1/2
As the rate is halved when the concentration of H2 is halved, the order of reaction with respect to H2 is 1.
Similarly, if we perform experiments while keeping the concentration of H2 constant and vary the concentration of NO:
Experiment 11:
[H2] = constant
[NO] = d
Experiment 12:
[H2] = constant
[NO] = 2d
(Rate 12) / (Rate 11) = (k[H2 ] [NO]) / (k[H2 ] [(2NO)])
= (k[H2 ] [NO]) / (k[H2 ] [2NO])
= 1/2
As the rate is halved when the concentration of NO is halved, the order of reaction with respect to NO is 1.
In summary,
Order of reaction with respect to H2 = 1
Order of reaction with respect to NO = 1
Therefore, the final order of reaction for each rate equation is:
1. Rate = k[H2] [NO]2 : First order with respect to H2, Second order with respect to NO.
2. Rate = k[H2]2 [O2] : Second order with respect to H2, First order with respect to O2.
3. Rate = k[H2 ][NO] : First order with respect to H2, First order with respect to NO.